1. Write the hyperbola in standard form

Given $16x^2-y^2+64x+4y+44=0$

Complete squares: $16(x^2+4x)-(y^2-4y)+44=0$ $16[(x+2)^2-4]-[(y-2)^2-4]+44=0$ $16(x+2)^2-64-(y-2)^2+4+44=0$ $16(x+2)^2-(y-2)^2-16=0$ $16(x+2)^2-(y-2)^2=16$ $\frac{(x+2)^2}{1}-\frac{(y-2)^2}{16}=1$

So the hyperbola has center $(-2,2)$.

• Transverse axis $T$ is the horizontal line through the center: $y=2$
• Conjugate axis $C$ is the vertical line through the center: $x=-2$

2. Interpret the required region

We need the region:

• above the parabola $x^2=y+4 \implies y=x^2-4,$
• below the transverse axis $y=2,$
• on the right of the conjugate axis $x=-2.$

Thus the region satisfies $x\ge -2, \qquad x^2-4\le y\le 2.$

For such a region to exist, $x^2-4\le 2 \implies x^2\le 6 \implies -\sqrt6\le x\le \sqrt6.$

Combining with $x\ge -2$, the valid interval is $-2\le x\le \sqrt6$ (since $-2>-\sqrt6$).

3. Set up the area integral

Area $A=\int_{-2}^{\sqrt6} \left[2-(x^2-4)\right]dx$ $A=\int_{-2}^{\sqrt6} (6-x^2)\,dx$

4. Evaluate the integral

$\int (6-x^2)dx=6x-\frac{x^3}{3}$

Hence $A=\left[6x-\frac{x^3}{3}\right]_{-2}^{\sqrt6}$

At $x=\sqrt6$: $6\sqrt6-\frac{(\sqrt6)^3}{3}=6\sqrt6-\frac{6\sqrt6}{3}=6\sqrt6-2\sqrt6=4\sqrt6$

At $x=-2$: $6(-2)-\frac{(-2)^3}{3}=-12+\frac{8}{3}=-\frac{28}{3}$

Therefore $A=4\sqrt6-\left(-\frac{28}{3}\right)=4\sqrt6+\frac{28}{3}$

5. Match with the options

This is $\boxed{4\sqrt6+\frac{28}{3}}$ which is Option C.

6. Compare with stored correct answer

Stored correct answer: C

Our derived answer is also C, so they agree.